University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 445: 21



Work Step by Step

$$A=\int^1_0\frac{dx}{(x+1)(x^2+1)}$$ 1) Express the integrand as a sum of partial fractions: Since $x^2+1$ is an irreducible quadratic factor, $$\frac{1}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$$ Clear fractions: $$Ax^2+A+Bx^2+(B+C)x+C=1$$ $$(A+B)x^2+(B+C)x+(A+C)=1$$ Equating coefficients of corresponding powers of $x$, we get - $A+B=0$ - $B+C=0$ - $A+C=1$ Solving for $A$, $B$, $C$, we get $A=1/2$, $B=-1/2$ and $C=1/2$. Therefore, $$\frac{1}{(x+1)(x^2+1)}=\frac{1}{2}\frac{1}{x+1}-\frac{1}{2}\frac{x-1}{x^2+1}$$ 2) Evaluate the integral: $$A=\frac{1}{2}\int^1_0\frac{dx}{x+1}-\frac{1}{2}\int^1_0\frac{x-1}{x^2+1}dx$$ $$A=\frac{1}{2}\ln|x+1|\Big]^1_0-\frac{1}{2}\Big(\int^1_0\frac{x}{x^2+1}dx-\int^1_0\frac{1}{x^2+1}dx\Big)$$ $$A=\frac{1}{2}\ln2-\frac{1}{2}\Big(\frac{1}{2}\int^1_0\frac{1}{x^2+1}d(x^2+1)-\tan^{-1}x\Big]^1_0\Big)$$ $$A=\frac{1}{2}\ln2-\frac{1}{2}\Big(\frac{1}{2}\ln|x^2+1|\Big]^1_0-\frac{\pi}{4}\Big)$$ $$A=\frac{1}{2}\ln2-\frac{1}{4}\ln2+\frac{\pi}{8}$$ $$A=\frac{1}{4}\ln2+\frac{\pi}{8}$$
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